`int_0^(1/2) arctanx/x dx` Use a power series to approximate the value of the integral with an error of less than 0.0001.

From the Power Series table for trigonometric function, we have:

`arctan(x) =sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)`

                     `= x -x^3/3 +x^5/5 - x^7/7 + x^9/9-...`

Applying it on the integral `int_0^(1/2) arctan(x)/x dx ` where the integrand is `f(x)=arctan(x)/x` , we get:

`int_0^(1/2) arctan(x)/x dx =int_0^1 arctan(x) *1/x dx`

                         `=int_0^(1/2)sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1) *1/xdx`

     ...

From the Power Series table for trigonometric function, we have:

`arctan(x) =sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)`

                     `= x -x^3/3 +x^5/5 - x^7/7 + x^9/9-...`

Applying it on the integral `int_0^(1/2) arctan(x)/x dx ` where the integrand is `f(x)=arctan(x)/x` , we get:

`int_0^(1/2) arctan(x)/x dx =int_0^1 arctan(x) *1/x dx`

                         `=int_0^(1/2)sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1) *1/xdx`

                         `=int_0^(1/2) sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1) *x^(-1)dx`

                         `=int_0^(1/2) sum_(n=0)^oo (-1)^n x^(2n+1-1)/(2n+1) dx`

                         `=int_0^(1/2) sum_(n=0)^oo (-1)^n x^(2n)/(2n+1) dx`

                         `=int_0^(1/2) [1 -x^2/3 +x^4/5 - x^6/7 +x^8/9-...] dx`

 Or 

`int_0^(1/2) arctan(x)/x dx =int_0^1 arctan(x) *1/x dx`

                         `=int_0^(1/2) 1/x* [x -x^3/3 +x^5/5 - x^7/7 +x^9/9-...] dx`

                         ` = int_0^(1/2) [x/x -x^3/(3x) +x^5/(5x) - x^7/(7x) +x^9/(9x)-...] dx`

                         `=int_0^(1/2) [1 -x^2/3 +x^4/5 - x^6/7 +x^8/9-...] dx`

To determine the indefinite integral, we integrate each term using Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` .

`int_0^(1/2) [1 -x^2/3+x^4/5 - x^6/7 +x^8/9-...] dx= [x -x^3/(3*3)+x^5/(5*5) - x^7/(7*7) +x^9/(9*9)-...]_0^1`

                                                    `=[x -x^3/9+x^5/25 - x^7/49 +x^9/81-...]_0^1`

Apply definite integral formula: `F(x)|a^b =F(b)-F(a)` .

`F(1/2) or F(0.5)=0.5 -0.5^3/9+0.5^5/25 - 0.5^7/49 +0.5^9/81-...`

                             `= 1/2 -1/72+1/800-1/6272+1/41472 -...`

`F(0)= 0 -0^3/9+0^5/25 - 0^7/49+0^9/81-...`

         `=0-0+0-0+0-...`

All the terms are 0 then `F(0)=0` .

We can stop on the 5th term  `(1/41472 ~~2.4113x10^(-5)) ` since we only need an error less than 0.0001.

`F(1/2)-F(0)= [1/2 -1/72+1/800-1/6272+1/41472]-[0]`

                       `=1/2 -1/72+1/800-1/6272+1/41472`

                       ` =0.487225785`

 Then, the approximated integral value will be:

`int_0^(1/2) arctan(x)/x dx~~0.4872`